Let $S={ mathcal{X_s} ;;|;; sinmathbb{R} }$. It is possible that $ Span(S) = mathbb{R}^{mathbb{R}}$












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Let $S={ mathcal{X_s} ;;|;; sinmathbb{R} }$.
It is possible that $ Span(S)=langle Srangle = mathbb{R}^{mathbb{R}}={f:mathbb{R} rightarrow mathbb{R} ; |;; f;; fuction }$?
I do not think this is possible, but I can not prove it. Can anyone give me a hint on how to prove this or a counter example. I also wanted a hint on how to describe the subspace generated $Span(S)=langle Srangle.$










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    1












    $begingroup$


    Let $S={ mathcal{X_s} ;;|;; sinmathbb{R} }$.
    It is possible that $ Span(S)=langle Srangle = mathbb{R}^{mathbb{R}}={f:mathbb{R} rightarrow mathbb{R} ; |;; f;; fuction }$?
    I do not think this is possible, but I can not prove it. Can anyone give me a hint on how to prove this or a counter example. I also wanted a hint on how to describe the subspace generated $Span(S)=langle Srangle.$










    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $S={ mathcal{X_s} ;;|;; sinmathbb{R} }$.
      It is possible that $ Span(S)=langle Srangle = mathbb{R}^{mathbb{R}}={f:mathbb{R} rightarrow mathbb{R} ; |;; f;; fuction }$?
      I do not think this is possible, but I can not prove it. Can anyone give me a hint on how to prove this or a counter example. I also wanted a hint on how to describe the subspace generated $Span(S)=langle Srangle.$










      share|cite|improve this question











      $endgroup$




      Let $S={ mathcal{X_s} ;;|;; sinmathbb{R} }$.
      It is possible that $ Span(S)=langle Srangle = mathbb{R}^{mathbb{R}}={f:mathbb{R} rightarrow mathbb{R} ; |;; f;; fuction }$?
      I do not think this is possible, but I can not prove it. Can anyone give me a hint on how to prove this or a counter example. I also wanted a hint on how to describe the subspace generated $Span(S)=langle Srangle.$







      linear-algebra vector-spaces






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      edited Jan 18 at 19:02









      José Carlos Santos

      164k22131234




      164k22131234










      asked Jan 18 at 17:40









      Ricardo FreireRicardo Freire

      513210




      513210






















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          $begingroup$

          I suppose that $chi_s(x)$ is $0$, unless $x=s$, in which case it is equal to $1$. If that's so, you are right, since every element of $operatorname{span}(S)$ is $0$ outside a finite set.






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          $endgroup$













          • $begingroup$
            Can you explain to me why outside a finite set $span (S) = { 0 }$?
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:00












          • $begingroup$
            I did not write that. Fortunately, since it doesn't make sense.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:02










          • $begingroup$
            Sorry, I got confused, I wanted to know if you have a way of describing the generated subspace $span(S)$
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:05






          • 1




            $begingroup$
            Sure: it's the space of those functions $fcolonmathbb{R}longrightarrowmathbb{R}$ such that the set ${xinmathbb{R},|,f(x)neq0}$ is finite.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:07











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          $begingroup$

          I suppose that $chi_s(x)$ is $0$, unless $x=s$, in which case it is equal to $1$. If that's so, you are right, since every element of $operatorname{span}(S)$ is $0$ outside a finite set.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you explain to me why outside a finite set $span (S) = { 0 }$?
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:00












          • $begingroup$
            I did not write that. Fortunately, since it doesn't make sense.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:02










          • $begingroup$
            Sorry, I got confused, I wanted to know if you have a way of describing the generated subspace $span(S)$
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:05






          • 1




            $begingroup$
            Sure: it's the space of those functions $fcolonmathbb{R}longrightarrowmathbb{R}$ such that the set ${xinmathbb{R},|,f(x)neq0}$ is finite.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:07
















          1












          $begingroup$

          I suppose that $chi_s(x)$ is $0$, unless $x=s$, in which case it is equal to $1$. If that's so, you are right, since every element of $operatorname{span}(S)$ is $0$ outside a finite set.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Can you explain to me why outside a finite set $span (S) = { 0 }$?
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:00












          • $begingroup$
            I did not write that. Fortunately, since it doesn't make sense.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:02










          • $begingroup$
            Sorry, I got confused, I wanted to know if you have a way of describing the generated subspace $span(S)$
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:05






          • 1




            $begingroup$
            Sure: it's the space of those functions $fcolonmathbb{R}longrightarrowmathbb{R}$ such that the set ${xinmathbb{R},|,f(x)neq0}$ is finite.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:07














          1












          1








          1





          $begingroup$

          I suppose that $chi_s(x)$ is $0$, unless $x=s$, in which case it is equal to $1$. If that's so, you are right, since every element of $operatorname{span}(S)$ is $0$ outside a finite set.






          share|cite|improve this answer









          $endgroup$



          I suppose that $chi_s(x)$ is $0$, unless $x=s$, in which case it is equal to $1$. If that's so, you are right, since every element of $operatorname{span}(S)$ is $0$ outside a finite set.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 18 at 17:55









          José Carlos SantosJosé Carlos Santos

          164k22131234




          164k22131234












          • $begingroup$
            Can you explain to me why outside a finite set $span (S) = { 0 }$?
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:00












          • $begingroup$
            I did not write that. Fortunately, since it doesn't make sense.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:02










          • $begingroup$
            Sorry, I got confused, I wanted to know if you have a way of describing the generated subspace $span(S)$
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:05






          • 1




            $begingroup$
            Sure: it's the space of those functions $fcolonmathbb{R}longrightarrowmathbb{R}$ such that the set ${xinmathbb{R},|,f(x)neq0}$ is finite.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:07


















          • $begingroup$
            Can you explain to me why outside a finite set $span (S) = { 0 }$?
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:00












          • $begingroup$
            I did not write that. Fortunately, since it doesn't make sense.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:02










          • $begingroup$
            Sorry, I got confused, I wanted to know if you have a way of describing the generated subspace $span(S)$
            $endgroup$
            – Ricardo Freire
            Jan 18 at 18:05






          • 1




            $begingroup$
            Sure: it's the space of those functions $fcolonmathbb{R}longrightarrowmathbb{R}$ such that the set ${xinmathbb{R},|,f(x)neq0}$ is finite.
            $endgroup$
            – José Carlos Santos
            Jan 18 at 18:07
















          $begingroup$
          Can you explain to me why outside a finite set $span (S) = { 0 }$?
          $endgroup$
          – Ricardo Freire
          Jan 18 at 18:00






          $begingroup$
          Can you explain to me why outside a finite set $span (S) = { 0 }$?
          $endgroup$
          – Ricardo Freire
          Jan 18 at 18:00














          $begingroup$
          I did not write that. Fortunately, since it doesn't make sense.
          $endgroup$
          – José Carlos Santos
          Jan 18 at 18:02




          $begingroup$
          I did not write that. Fortunately, since it doesn't make sense.
          $endgroup$
          – José Carlos Santos
          Jan 18 at 18:02












          $begingroup$
          Sorry, I got confused, I wanted to know if you have a way of describing the generated subspace $span(S)$
          $endgroup$
          – Ricardo Freire
          Jan 18 at 18:05




          $begingroup$
          Sorry, I got confused, I wanted to know if you have a way of describing the generated subspace $span(S)$
          $endgroup$
          – Ricardo Freire
          Jan 18 at 18:05




          1




          1




          $begingroup$
          Sure: it's the space of those functions $fcolonmathbb{R}longrightarrowmathbb{R}$ such that the set ${xinmathbb{R},|,f(x)neq0}$ is finite.
          $endgroup$
          – José Carlos Santos
          Jan 18 at 18:07




          $begingroup$
          Sure: it's the space of those functions $fcolonmathbb{R}longrightarrowmathbb{R}$ such that the set ${xinmathbb{R},|,f(x)neq0}$ is finite.
          $endgroup$
          – José Carlos Santos
          Jan 18 at 18:07


















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